Question: 17. The sequence (Xi ) of independent, identically distributed random variables is such that P(Xi = 0) = 1 p, P(Xi = 1) =

17. The sequence (Xi ) of independent, identically distributed random variables is such that P(Xi = 0) = 1 − p, P(Xi = 1) = p.

If f is a continuous function on [0, 1], prove that Bn(p) = E



f



X1 + · · · + Xn n



152 The main limit theorems is a polynomial in p of degree at most n. Use Chebyshev’s inequality to prove that for all p with 0 ≤ p ≤ 1, and any ǫ > 0, X k∈K



n k



pk (1 − p)n−k ≤

1 4nǫ2

, where K = {k : 0 ≤ k ≤ n, |k/n − p| > ǫ}. Using this and the fact that f is bounded and uniformly continuous in [0, 1], prove the following version of the Weierstrass approximation theorem:

lim n→∞

sup 0≤p≤1 | f (p) − Bn(p)| = 0.

(Oxford 1976F)

Step by Step Solution

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Students Have Also Explored These Related Elementary Probability For Applications Questions!